受Raymond Chen帖子的启发，假设你有一个4x4二维数组，写一个函数使它旋转90度。Raymond链接到伪代码中的解决方案，但我想看到一些真实的东西。
[1][2][3][4]
[5][6][7][8]
[9][0][1][2]
[3][4][5][6]
就变成:
[3][9][5][1]
[4][0][6][2]
[5][1][7][3]
[6][2][8][4]
更新:Nick的答案是最直接的，但是有没有比n²更好的方法呢?如果矩阵是10000x10000呢?
当前回答
下面是一个c#静态泛型方法，它可以为您完成这项工作。变量的名称很好，所以您可以很容易地理解算法的思想。
private static T[,] Rotate180 <T> (T[,] matrix)
{
var height = matrix.GetLength (0);
var width = matrix.GetLength (1);
var answer = new T[height, width];
for (int y = 0; y < height / 2; y++)
{
int topY = y;
int bottomY = height - 1 - y;
for (int topX = 0; topX < width; topX++)
{
var bottomX = width - topX - 1;
answer[topY, topX] = matrix[bottomY, bottomX];
answer[bottomY, bottomX] = matrix[topY, topX];
}
}
if (height % 2 == 0)
return answer;
var centerY = height / 2;
for (int leftX = 0; leftX < Mathf.CeilToInt(width / 2f); leftX++)
{
var rightX = width - 1 - leftX;
answer[centerY, leftX] = matrix[centerY, rightX];
answer[centerY, rightX] = matrix[centerY, leftX];
}
return answer;
}
其他回答
我的c#示例代码的伟大算法发送@dimple:
/* Author: Dudi,
* http://www.tutorialspoint.com/compile_csharp_online.php?PID=0Bw_CjBb95KQMYm5qU3VjVGNuZFU */
using System.IO;
using System;
class Program
{
static void Main()
{
Console.WriteLine("Rotating this matrix by 90+ degree:");
int[,] values=new int[3,3]{{1,2,3}, {4,5,6}, {7,8,9}};
//int[,] values=new int[4,4]{{101,102,103, 104}, {105,106, 107,108}, {109, 110, 111, 112}, {113, 114, 115, 116}};
print2dArray(ref values);
transpose2dArray(ref values);
//print2dArray(ref values);
reverse2dArray(ref values);
Console.WriteLine("Output:");
print2dArray(ref values);
}
static void print2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
Console.Write(matrix[n,m] +"\t");
}
Console.WriteLine();
}
Console.WriteLine();
}
static void transpose2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen; m++){
if(n>m){
int tmp = matrix[n,m];
matrix[n,m] = matrix[m,n];
matrix[m,n] = tmp;
}
}
}
}
static void reverse2dArray(ref int[,] matrix){
int nLen = matrix.GetLength(0);
int mLen = matrix.GetLength(1);
for(int n=0; n<nLen; n++){
for(int m=0; m<mLen/2; m++){
int tmp = matrix[n,m];
matrix[n,m] = matrix[n, mLen-1-m];
matrix[n,mLen-1-m] = tmp;
}
}
}
}
/*
Rotating this matrix by 90+ degree:
1 2 3
4 5 6
7 8 9
Output:
7 4 1
8 5 2
9 6 3
*/
Javascript解决NxN矩阵与运行时O(N^2)和内存O(1)
function rotate90(matrix){
var length = matrix.length
for(var row = 0; row < (length / 2); row++){
for(var col = row; col < ( length - 1 - row); col++){
var tmpVal = matrix[row][col];
for(var i = 0; i < 4; i++){
var rowSwap = col;
var colSwap = (length - 1) - row;
var poppedVal = matrix[rowSwap][colSwap];
matrix[rowSwap][colSwap] = tmpVal;
tmpVal = poppedVal;
col = colSwap;
row = rowSwap;
}
}
}
}
对于i:= 0到X 对于j:= 0到X做 [j][i]:= graphic2[X-i][j]
X是图形所在数组的大小。
这是我对矩阵90度旋转的尝试，这是c中的2步解决方案，首先转置矩阵，然后交换cols。
#define ROWS 5
#define COLS 5
void print_matrix_b(int B[][COLS], int rows, int cols)
{
for (int i = 0; i <= rows; i++) {
for (int j = 0; j <=cols; j++) {
printf("%d ", B[i][j]);
}
printf("\n");
}
}
void swap_columns(int B[][COLS], int l, int r, int rows)
{
int tmp;
for (int i = 0; i <= rows; i++) {
tmp = B[i][l];
B[i][l] = B[i][r];
B[i][r] = tmp;
}
}
void matrix_2d_rotation(int B[][COLS], int rows, int cols)
{
int tmp;
// Transpose the matrix first
for (int i = 0; i <= rows; i++) {
for (int j = i; j <=cols; j++) {
tmp = B[i][j];
B[i][j] = B[j][i];
B[j][i] = tmp;
}
}
// Swap the first and last col and continue until
// the middle.
for (int i = 0; i < (cols / 2); i++)
swap_columns(B, i, cols - i, rows);
}
int _tmain(int argc, _TCHAR* argv[])
{
int B[ROWS][COLS] = {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20},
{21, 22, 23, 24, 25}
};
matrix_2d_rotation(B, ROWS - 1, COLS - 1);
print_matrix_b(B, ROWS - 1, COLS -1);
return 0;
}
Python:
rotated = list(zip(*original[::-1]))
和逆时针方向:
rotated_ccw = list(zip(*original))[::-1]
这是如何工作的:
Zip (*original)将通过将列表中的对应项堆叠到新的列表中来交换2d数组的轴。(*操作符告诉函数将包含的列表分布到参数中)
>>> list(zip(*[[1,2,3],[4,5,6],[7,8,9]]))
[[1,4,7],[2,5,8],[3,6,9]]
语句[::-1]反转数组元素(请参阅扩展切片或这个问题):
>>> [[1,2,3],[4,5,6],[7,8,9]][::-1]
[[7,8,9],[4,5,6],[1,2,3]]
最后，将两者结合就得到了旋转变换。
改变[::-1]的位置将使列表在矩阵的不同层次上颠倒。
